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STAT3801_2011Unit4b

# STAT3801_2011Unit4b - THE UNIVERSITY OF HONG KONG...

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THE UNIVERSITY OF HONG KONG DEPARTMENT OF STATISTICS AND ACTUARIAL SCIENCE STAT3801 ADVANCED LIFE CONTINGENCIES Unit 4b 2010 - 11 2 nd semester 4.14 Valuation of benefits in multiple decrement context Discrete life insurance for decrement j APV = X k =0 b k +1 · v k +1 · k | q ( j ) x Continuous life insurance for decrement j APV = Z 0 b t e - δt · t p ( τ ) x μ ( j ) x ( t ) dt Life annuity-due ¨ a ( τ ) x = X k =0 1 · v k k p ( τ ) x Actuarial present values and numerical evaluation A = r X j =1 Z 0 B ( j ) x + t v t t p ( τ ) x μ ( j ) x ( t ) dt Double indemnity provision for n - year term insurance A = 2 Z n 0 v t t p ( τ ) x μ (1) x ( t ) dt + Z n 0 v t t p ( τ ) x μ (2) x ( t ) dt

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S&AS: STAT3801 ADVANCED LIFE CONTINGENCIES 2 Numerical evaluation assuming UDD in multiple decrement model 2 Z n 0 v t t p ( τ ) x μ (1) x ( t ) dt = 2 n - 1 X k =0 v k k p ( τ ) x Z 1 0 v s s p ( τ ) x + k μ (1) x ( k + s ) ds = 2 n - 1 X k =0 v k +1 k p ( τ ) x q (1) x + k Z 1 0 (1 + i ) 1 - s ds = 2 i δ n - 1 X k =0 v k +1 k p ( τ ) x q (1) x + k Applying similar argument for second integral and combining A = i δ ( n - 1 X k =0 v k +1 k p ( τ ) x h 2 q (1) x + k + q (2) x + k i ) = i δ n - 1 X k =0 v k +1 k p ( τ ) x h q (1) x + k + q ( τ ) x + k i = A (1) 1 x : n + A 1 x : n
S&AS: STAT3801 ADVANCED LIFE CONTINGENCIES 3 If accidental death benefit = t and no other death benefits A = Z 0 t v t t p ( τ ) x μ (1) x ( t ) dt = X k =0 v k k p ( τ ) x Z 1 0 ( k + s ) v s s p ( τ ) x + k μ (1) x ( k + s ) ds = X k =0 v k +1 k p ( τ ) x q (1) x + k Z 1 0 ( k + s )(1 + i ) 1 - s ds = X k =0 v k +1 k p ( τ ) x q (1) x + k i δ k + 1 δ - 1 i Using midpoint rule for approximate integration Z 1 0 ( k + s )(1 + i ) 1 - s ds k + 1 2 v - 1 2 A = X k =0 v k + 1 2 k p ( τ ) x q (1) x + k k + 1 2 For varying benefits assuming UDD of j th decrement in multiple decrement model X k =0 v k +1 k p ( τ ) x q ( j ) x + k Z 1 0 B ( j ) x + k + s (1 + i ) 1 - s ds = X k =0 v k + 1 2 k p ( τ ) x q ( j ) x + k B ( j ) x + k + 1 2 by midpoint integration rule.

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S&AS: STAT3801 ADVANCED LIFE CONTINGENCIES 4 Example 4.9: In single decrement model for decrement 3, 1 2 at midyear and 1 2 at year end, UDD for other single decrements 1 2 q 0 (1) x + k 1 2 q 0 (1) x + k 1 2 q 0 (2) x + k 1 2 q 0 (3) x + k 1 2 q 0 (2) x + k 1 2 q 0 (3) x + k - - - - - - -- - - - - - - -- | | | | | time 0 1 4 1 2 3 4 1 age x + k x + k + 1 4 x + k + 1 2 x + k + 3 4 x + k + 1 Actuarial present value A = X k =0 v k k p ( τ ) x h ( 1 2 + ) q (3) x + k v 1 2 B (3) x + k + 1 2 + v 1 2 ( 1 2 + ) p ( τ ) x + k · ( 1 2 + ) q (3) x + k + ( 1 2 + ) v 1 2 B (3) x + k +1 Now ( 1 2 + ) q (3) x + k = ( 1 2 - ) p ( τ ) x + k 1 2 q 0 (3) x + k = 1 2 p 0 (1) x + k · 1 2 p 0 (2) x + k 1 2 q 0 (3) x + k ( 1 2 + ) p ( τ ) x + k = 1 2 p 0 (1) x + k · 1 2 p 0 (2) x + k 1 - 1 2 q 0 (3) x + k Note The question states that for each survivor at time 0 in the
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